Is x < 0?

[Max@Math Revolution]

[GMAT math practice question]

Is x < 0?

1) x^3 + 1 < 0
2) x^3 + x + 1 < 0

[Brent@GMATPrepNow]

[GMAT math practice question]

Is x < 0?

1) x³ + 1 < 0
2) x³ + x + 1 < 0

Target question: Is x < 0?

Two important rules:
ODD exponents preserve the sign of the base.
So, (NEGATIVE)^(ODD integer) = NEGATIVE
and (POSITIVE)^(ODD integer) = POSITIVE

An EVEN exponent always yields a positive result (unless the base = 0)
So, (NEGATIVE)^(EVEN integer) = POSITIVE
and (POSITIVE)^(EVEN integer) = POSITIVE

Statement 1: x³ + 1 < 0
Subtract 1 from both sides to get: x³ < -1
If x³ is less than -1, then x³ is NEGATIVE
Since an odd power (like 3) preserves the sign of the base, we know that x is NEGATIVE
The answer to the target question is YES, it is true that x < 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³ + x + 1 < 0
Subtract 1 from both sides to get: x³ + x < -1
Let's see what happens when x is POSITIVE, and when x is NEGATIVE

If x is POSITIVE, we get: POSITIVE³ + POSITIVE < -1
Use the above rule to simplify: POSITIVE + POSITIVE < -1
This is IMPOSSIBLE, so it CANNOT be the case that x is positive

If x is NEGATIVE, we get: NEGATIVE³ + NEGATIVE< -1
Use the above rule to simplify: NEGATIVE + NEGATIVE < -1
PERFECT!
The answer to the target question is YES, it is true that x < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

[GMATGuruNY]

[GMAT math practice question]

Is x < 0?

1) x^3 + 1 < 0
2) x^3 + x + 1 < 0

Statement 1:
x³ < -1
The cube of x will be negative only if x itself is negative.
Thus, x < 0.
SUFFICIENT.

Statement 2:
x³ + x < -1
x(x² + 1) < -1
Since the square of a value cannot be negative, x² ≥ 0, with the result that the factor in blue must be POSITIVE.
Implication:
For the left side to yield a negative product, the factor in red must be NEGATIVE.
Thus, x < 0.
SUFFICIENT.

The correct answer is D.

[Max@Math Revolution]

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
x^3 + 1 < 0
=> (x+1)(x^2-x+1) < 0
=> x + 1 < 0 since x^2-x+1 > 0
=> x < -1 < 0
Thus, condition 1) is sufficient, and the answer is 'yes'.

Condition 2)
x^3 + x + 1 < 0
=> x^3 + x < -1
=> x(x^2 + 1) < -1
=> x < -1/(x^2 + 1) since x^2 + 1 > 0
=> x < -1/(x^2 + 1) < 0 since x^2 + 1 > 0
Thus, condition 2) is sufficient, and the answer is 'yes'.

Therefore, D is the answer.
Answer: D